Print Grid Solutions

print_grid.rb

# This is a conglomeration of solution possibilites, collected and put together by Ryan Matthews - reposted here by Timothy

def print_grid(x=5,y=5)
  size = 4
  0.upto(size) do |a|
    0.upto(size) do |b|
      if b == x && a == y
        print "X"
      else
        print "0"
      end
    end
    puts ""
  end
  nil
end

# Next! This will be an excercise for you to walk through. You might find it helpful to try each line individually in irb.


def print_grid(x=5,y=5)
  matrix = Array.new(5) { Array.new(5,0) }
  matrix[y][x] = "X" if matrix[y] && matrix[y][x]
  matrix.each { |r| puts r.join }
  nil
end

# The most complex part here is the first line. We are creating a new Array, telling it we want 5 elements, and we pass a block. That block will set the value for each element in the array. In this case we are creating a new array with 5 elements, each of which is a 0.


def print_grid(x=nil,y=nil)
  grid = Array.new(5) {"O" * 5}
  grid[x][y] = "X" if x && y
  puts grid
end

# This is a mix of the previous two code examples. Much cleaner! Refactoring is fun.

# This is just showing off. Getting the method down to 2 lines.

require 'matrix'
def print_grid(x=nil, y=nil)
  m = Matrix.build(5,5) { |r,c| (r==y && c==x)? "X" : "0" }
  m.to_a.map {|i| puts i.join}
end

# The most important thing to understand here is the turnary. That is the if/else statement on the first line.
# 
# ```ruby
# >> (true) ? "hi" : "bye"
# => "hi"
# >> (false) ? "hi" : "bye"
# => "bye"
# ```
# 
# You don't need to understand the Matrix stuff. If you want to look into it, go for it.